1.
Warm up
A. Find the average rate of change on [1, 2]
of the function: f (x) = x3 - 2x
B. Find the instantaneous rate of change when
x = 2 of the same function.
2.
The Lame Joke of the day..
And now it’s time for..
Why did the scientist install a knocker on
his door instead of a door bell?
He wanted to win the No-bell prize
3.
The Lame Joke of the day..
And now it’s time for..
What is the formed by Ba + Na 2?
Banana
4.
The Mean Value Theorem
Lesson 4.2
I wonder how
mean this
theorem really
is?
6.
Think About It
• Consider a trip of two hours
that is 120 miles in distance …
You have averaged 60 miles per hour
• What reading on your speedometer would
you have expected to see at least once?
60
7.
Rolle’s Theorem
• Given f(x) on closed interval [a, b]
Differentiable on open interval (a, b)
• If f(a) = f(b) … then
There exists at least one number
a < c < b such that f ’(c) = 0
f(a) = f(b)
a bc
8.
Mean Value Theorem
• We can “tilt” the picture of Rolle’s Theorem
Stipulating that f(a) ≠ f(b)
• Then there exists
a c such that
a b
c
( ) ( )
'( )
f b f a
f c
b a
9.
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b f a
f c
b a
Mean Value Theorem for Derivatives
10.
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b f a
f c
b a
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.
11.
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b f a
f c
b a
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.
The Mean Value Theorem only applies over a closed interval.
12.
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b f a
f c
b a
Mean Value Theorem for Derivatives
The Mean Value Theorem says that at some point
in the closed interval, the actual slope
(instantaneous rate of change) equals the
average slope (average rate of change).
13.
y
x
0
A
B
a b
Slope of chord:
f b f a
b a
Slope of tangent:
f c
y f x
Tangent parallel
to chord.
c
14.
Example
• Suppose you have a function f(x) = x^2 in
the interval [1, 3].
ab
afbf
cf
)()(
)('cnumberaexiststherestatesMVTD
4
2
8
13
)1()3(
:changeofrateaverage
ff
To find c,
set
derivative =
to 4
xxf 2)('
42x
2x
15.
2][-2,intervalon the12)(functionheConsider t 3
xxxf
ab
afbf
cf
)()(
)('cnumberastatesMVTD
8
22
)2()2( ff
average
123)(' 2
xxf
8123)(' 2
ccf
3
42
c
155.1
3
2
c
There can
be more
than 1
value of c
that satisfy
MVTD
16.
2][-2,intervalon the
1
)(functionheConsider t
x
xf
ab
afbf
cf
)()(
)('cnumberastatesMVTD
There is no value of
c that satisfies this
equation. Why?
f(x) is not
continuous in the
given interval!
17.
17
Speeding problem
Two police cars are located at fixed points 5
miles apart on a long straight road.
• The speed limit is 55 mph
• A car passes the first point at 53 mph
• Four minutes later he passes the second car at
48 mph
I think he was
speeding
We need to
prove it
Can the officers issue the man a ticket?
18.
Given f(x) = 5 – 4/x, find all c in the interval (1,4)
such that the slope of the secant line = the slope of
the tangent line.
?
14
)1()4(
)('
ff
cf 1
14
14
?)(' xf 2
4
x
1
4
2
x
2
4 x
2x
But in the interval of (1,4),
only 2 works, so c = 2.
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